Question

The vehicle starting from rest is accelerated along straight line where acceleration is governed by relation, a=(10/2v+3), where 'a' is the acceleration in m / sec² and v is the velocity in m/ sec. Determine 1.time taken by vehicle to reach velocity of 12 m/ sec. 2.distance moved during that time.​

Answer 1

Explanation:

Initial velocity, u = 0 m/sec

Final velocity, v = 12 m/sec

acceleration, a = 10/(2*12) + 3 = 5/12 + 3 = 41/12 m/sec²

1.) formula

v= u + at

12 = 0 + 41/12 × t

t = 144/41 secs

2.) formula

v² - u² = 2as

12² - 0² = 2 × 41/12 × s

s= 864/41 m

[Note: if the equation for acceleration is

(10/2)*v + 3 instead of (10/(2v) + 3 then

a= 63 , t = 4/21 secs , s = 8/7 m ]

Answer 2

Explanation:

see please the attached document for the detailed explanation and step by step working.

t = 1/5 × Ln 21

S = 63/5 - 1.5 Ln 21